Visit the website 123iitjee.manishverma.site for latest posts, courses, admission & more.

### The half-life of a radioactive nuclide is 100 hours ....

The half-life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be:

(1) $\frac {1}{2}$                    (2) $\frac {1}{2\sqrt 2}$
(3) $\frac {2}{3}$                    (4) $\frac {2}{3\sqrt 2}$

Solution

Activity $A \propto N = \frac {N_0}{2^n}$

$A_0 \propto N_0$

Fraction of original activity $\frac {A}{A_0}=\frac {1}{2^n}$, where n = number of half-lives

If half-life is 100 hours, number of half-lives in 150 hours = 1.5 = n

So, $\frac {A}{A_0}=\frac {1}{2^{1.5}}=\frac {1}{2\sqrt 2}$

### Sum of the coefficients in the expansion of $(x+y)^n$ ....
If the sum of the coefficients in the expansion of $(x+y)^n$ is 4096, then the greatest coefficient in the expansion is _ _ _ _ . Solution $C_0 + C_1 + C_2 + C_3 + ......................... + C_n =4096$ $\therefore 2^n = 4096 =2^{12}$ $\Rightarrow n = 12$ Greatest coefficient = ${}^{12}{C_6} = 924$